Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

TOP1(mark1(X)) -> PROPER1(X)
PROPER1(g1(X)) -> G1(proper1(X))
PROPER1(g1(X)) -> PROPER1(X)
PROPER1(f1(X)) -> PROPER1(X)
TOP1(ok1(X)) -> TOP1(active1(X))
ACTIVE1(c) -> F1(g1(c))
F1(ok1(X)) -> F1(X)
TOP1(mark1(X)) -> TOP1(proper1(X))
G1(ok1(X)) -> G1(X)
ACTIVE1(c) -> G1(c)
TOP1(ok1(X)) -> ACTIVE1(X)
PROPER1(f1(X)) -> F1(proper1(X))

The TRS R consists of the following rules:

active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

TOP1(mark1(X)) -> PROPER1(X)
PROPER1(g1(X)) -> G1(proper1(X))
PROPER1(g1(X)) -> PROPER1(X)
PROPER1(f1(X)) -> PROPER1(X)
TOP1(ok1(X)) -> TOP1(active1(X))
ACTIVE1(c) -> F1(g1(c))
F1(ok1(X)) -> F1(X)
TOP1(mark1(X)) -> TOP1(proper1(X))
G1(ok1(X)) -> G1(X)
ACTIVE1(c) -> G1(c)
TOP1(ok1(X)) -> ACTIVE1(X)
PROPER1(f1(X)) -> F1(proper1(X))

The TRS R consists of the following rules:

active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 4 SCCs with 6 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

G1(ok1(X)) -> G1(X)

The TRS R consists of the following rules:

active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


G1(ok1(X)) -> G1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(G1(x1)) = 2·x1   
POL(ok1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

F1(ok1(X)) -> F1(X)

The TRS R consists of the following rules:

active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


F1(ok1(X)) -> F1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(F1(x1)) = 2·x1   
POL(ok1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

PROPER1(g1(X)) -> PROPER1(X)
PROPER1(f1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


PROPER1(f1(X)) -> PROPER1(X)
The remaining pairs can at least be oriented weakly.

PROPER1(g1(X)) -> PROPER1(X)
Used ordering: Polynomial interpretation [21]:

POL(PROPER1(x1)) = 2·x1   
POL(f1(x1)) = 2 + 2·x1   
POL(g1(x1)) = 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
          ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

PROPER1(g1(X)) -> PROPER1(X)

The TRS R consists of the following rules:

active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


PROPER1(g1(X)) -> PROPER1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(PROPER1(x1)) = 2·x1   
POL(g1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

TOP1(ok1(X)) -> TOP1(active1(X))
TOP1(mark1(X)) -> TOP1(proper1(X))

The TRS R consists of the following rules:

active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


TOP1(mark1(X)) -> TOP1(proper1(X))
The remaining pairs can at least be oriented weakly.

TOP1(ok1(X)) -> TOP1(active1(X))
Used ordering: Polynomial interpretation [21]:

POL(TOP1(x1)) = 2·x1   
POL(active1(x1)) = x1   
POL(c) = 3   
POL(f1(x1)) = 2·x1   
POL(g1(x1)) = 1   
POL(mark1(x1)) = 1 + x1   
POL(ok1(x1)) = x1   
POL(proper1(x1)) = x1   

The following usable rules [14] were oriented:

proper1(f1(X)) -> f1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(g1(X)) -> g1(proper1(X))
g1(ok1(X)) -> ok1(g1(X))
proper1(c) -> ok1(c)
active1(c) -> mark1(f1(g1(c)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

TOP1(ok1(X)) -> TOP1(active1(X))

The TRS R consists of the following rules:

active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


TOP1(ok1(X)) -> TOP1(active1(X))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(TOP1(x1)) = 2·x1   
POL(active1(x1)) = 0   
POL(c) = 0   
POL(f1(x1)) = 0   
POL(g1(x1)) = 0   
POL(mark1(x1)) = 0   
POL(ok1(x1)) = 2   

The following usable rules [14] were oriented:

active1(f1(g1(X))) -> mark1(g1(X))
active1(c) -> mark1(f1(g1(c)))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

active1(c) -> mark1(f1(g1(c)))
active1(f1(g1(X))) -> mark1(g1(X))
proper1(c) -> ok1(c)
proper1(f1(X)) -> f1(proper1(X))
proper1(g1(X)) -> g1(proper1(X))
f1(ok1(X)) -> ok1(f1(X))
g1(ok1(X)) -> ok1(g1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.